Jan 21, 2018 · tan^(-1) 1 We find: tan 1 ~~ 1.5574 tan^(-1) 1 = pi/2 ~~ 1.5757 So tan^(-1) 1 is slightly greater than tan 1

If you have additional information, regarding which quadrant, you should to one of the two following computations: For the second quadrant: theta = tan^-1(-1) + pi theta = -pi/4 + pi theta …

May 27, 2016 · Let #sin^-1x=theta# hence #x=sintheta#. For #0<x<1# we draw a right triangle with hypotenuse equal to 1 and the other side equals to #x# like the one in the Figure below.

Jan 4, 2017 · How do you use inverse trigonometric functions to find the solutions of the equation that are in

Feb 27, 2018 · 794 views around the world You can reuse this answer ...

Aug 31, 2016 · 11pi/6 say x = tan^-1 tan(11pi/6) so, tan x = tan(11pi/6) or, x = 11pi/6

Sep 7, 2015 · (3pi)/4 radians tan(pi-pi/4)=tan(3pi/4) = -1. Hence arctan(-1)=(3pi)/4 radians

May 13, 2016 · +-3/sqrt 13 Let a = tan^(-1)(2/3). Tangent is positive in both the first and the third quadrants. The given expression is cos a = +-3/sqrt 13. Positive sign for a in the first quadrant …

Apr 19, 2016 · Let #a =tan^(-1)(2/3)#. As tan a = 2/3, cos a and sin a should both have the same sign and #a = tan^(-1)(2/3)=33.7^o or 213.7^o#, in #[0, 2pi]#.

Oct 12, 2016 · The restriction for #tan^-1x # is #(-pi/2,pi/2)# but #(11pi)/10# is in quadrant III. Since tangent is positive in quadrant III our answer will be in the positive quadrant from the …